A projectile fired from the point​ (0,0) at an angle to the positive​ x-axis has a trajectory given by y equals Cx minus left parenthesis 1 plus Upper C squared right parenthesis left parenthesis StartFraction g Over 2 EndFraction right parenthesis left parenthesis StartFraction x Over v EndFraction right parenthesis squared .
In this​ formula, x is the horizontal distance in​ meters, y is the height in​ meters, v is the initial velocity in meters per​ second, gequals9.81 m divided by sec squared is the acceleration due to​ gravity, and Upper C greater than 0 is a constant determined by the angle of elevation.
A howitzer fires an artillery round with a velocity of 881 ​m/sec. Answer parts​ (a) and​ (b).
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Part 1
​(a) If the round must clear a hill 223 meters high at a distance of 2427 meters in front of the​ howitzer, what C values are permitted in the trajectory​ equation?

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The equation for the trajectory of the projectile is given by:

\[
y = Cx - \left(1 + C^2 \right) \left( \frac{g}{2} \right) \left( \frac{x^2}{v^2} \right)
\]

Where:
- \( y \) is the height of the projectile at a given horizontal distance \( x \),
- \( C \) is a constant determined by the angle of elevation,
- \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity,
- \( v = 881 \, \text{m/s} \) is the initial velocity of the projectile,
- \( x = 2427 \, \text{m} \) is the horizontal distance at which the projectile must clear the hill,
- \( y = 223 \, \text{m} \) is the height of the hill.

### Step 1: Plugging values into the trajectory equation

We are looking for the value of \( C \) that satisfies the given conditions. So, we will substitute \( y = 223 \, \text{m} \), \( x = 2427 \, \text{m} \), \( v = 881 \, \text{m/s} \), and \( g = 9.81 \, \text{m/s}^2 \) into the equation:

\[
223 = C \cdot 2427 - \left( 1 + C^2 \right) \left( \frac{9.81}{2} \right) \left( \frac{2427^2}{881^2} \right)
\]

### Step 2: Simplify the equation

The term involving gravity and velocity can be simplified:

\[
\left( \frac{9.81}{2} \right) \left( \frac{2427^2}{881^2} \right)
\]

Let's calculate this term:

\[
\frac{9.81}{2} = 4.905
\]
\[
\frac{2427^2}{881^2} = \left( \frac{2427}{881} \right)^2 = 7.577
\]
\[
4.905 \times 7.577 = 37.158
\]

So the equation becomes:

\[
223 = 2427C - (1 + C^2) \times 37.158
\]

### Step 3: Rearrange the equation

Expanding the right-hand side:

\[
223 = 2427C - 37.158 - 37.158C^2
\]

Move all terms to one side to form a quadratic equation in \( C \):

\[
37.158C^2 - 2427C + 260.158 = 0
\]

### Step 4: Solve the quadratic equation

We can now solve the quadratic equation using the quadratic formula:

\[
C = \frac{-(-2427) \pm \sqrt{(-2427)^2 - 4 \times 37.158 \times 260.158}}{2 \times 37.158}
\]

Let's calculate the discriminant and solve for \( C \):

\[
C = \frac{2427 \pm \sqrt{2427^2 - 4 \times 37.158 \times 260.158}}{2 \times 37.158}
\]

Now, calculating each term:

\[
2427^2 = 5890929
\]
\[
4 \times 37.158 \times 260.158 = 38662.24
\]
\[
2427^2 - 38662.24 = 5852266.76
\]
\[
\sqrt{5852266.76} = 2419.19
\]

So, the two possible values for \( C \) are:

\[
C = \frac{2427 \pm 2419.19}{74.316}
\]

This gives two solutions:

1. \( C = \frac{2427 + 2419.19}{74.316} = \frac{4846.19}{74.316} \approx 65.19 \)
2. \( C = \frac{2427 - 2419.19}{74.316} = \frac{7.81}{74.316} \approx 0.105 \)

### Final Answer:

The permitted values for \( C \) are approximately **65.19** and **0.105**.

---

Would you like further details on solving quadratic equations or any other part of the solution?

Here are some additional questions:
1. How do we interpret the values of \( C \) in terms of angle of elevation?
2. How would the values of \( C \) change if the hill were taller?
3. What happens to the trajectory if the initial velocity is increased?
4. How do wind or air resistance affect the values of \( C \)?
5. How is the time of flight affected by different values of \( C \)?

**Tip**: Solving quadratic equations often gives two possible answers, but it's important to check which one makes sense in the context of the problem.

A projectile fired from the point​ (0,0) at an angle to the positive​ x-axis has a trajectory given by y = Cx - (1 + C^2)(g/2)(x/v)^2. In this formula, x is the horizontal distance in meters, y is the height in meters, v is the initial velocity in meters per second, g = 9.81 m/s² is the acceleration due to gravity, and C > 0 is a constant determined by the angle of elevation. A howitzer fires an artillery round with a velocity of 881 m/sec. If the round must clear a hill 223 meters high at a distance of 2427 meters in front of the howitzer, what C values are permitted in the trajectory equation?

Learn how to solve for C in a projectile motion problem where a howitzer fires a round with a velocity of 881 m/s. The round must clear a hill 223 meters high, located 2427 meters away. This problem involves algebraic manipulation and solving quadratic equations, suitable for undergraduate-level students.

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